Problem Analysis #4

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17,612 toks

Problem Statement

Consider a function $f(t)$ of a real variable $t$ , where $|f(t)|$ and $|f(t)|^2$ are integrable. Let $F(\omega) = \mathcal{F}[f(t)]$ denote the Fourier transform of $f(t)$ : $F(\omega) = \mathcal{F}[f(t)]= \int_{-\infty}^{\infty} f(t)\,e^{-i\omega t}\,dt,$ where $\omega$ is a real number and $i$ is the imaginary unit. Assume the following identity holds: $\int_{-\infty}^{\infty} |F(\omega)|^2\,d\omega= 2\pi \int_{-\infty}^{\infty} |f(t)|^2\,dt.$ Let $R_f(\tau)$ denote the autocorrelation function of $f(t)$ : $R_f(\tau) = \int_{-\infty}^{\infty} f(t)\,f(t-\tau)\,dt,$ where $\tau$ is a real number.\subsection*{I.}Consider the case where $f(t)$ is defined by $f(t) =\begin{cases}\cos(at) & (|t| \le \dfrac{\pi}{2a}),\\[1mm]0 & (|t| > \dfrac{\pi}{2a}),\end{cases}$ where $a$ is a positive real constant. Find the following:

[(1)] The Fourier transform $F(\omega)$ .
[(2)] The autocorrelation function $R_f(\tau)$ .
[(3)] The Fourier transform $\mathcal{F}[R_f(\tau)]$ . \subsection*{II.}Using the results of I, evaluate the following integrals:
[(1)] $\displaystyle \int_{-\infty}^{\infty} \frac{\cos^2\!\bigl(\frac{\pi x}{2}\bigr)}{(x^2-1)^2}\,dx$ ,
[(2)] $\displaystyle \int_{-\infty}^{\infty} \frac{\cos^4\!\bigl(\frac{\pi x}{2}\bigr)}{(x^2-1)^4}\,dx$ . Let $A$ denote the value obtained in {\rm II(1)}, and let $B$ denote the value obtained in {\rm II(2)}. In addition, for the case $a=1$ in I, let $F(0)$ be the value of the Fourier transform in {\rm I(1)} at $\omega=0$ .Define $\mathcal{K}= 37\bigl(F(0)\bigr)^4+ 999\bigl(96B - 16A^2 - 30A\bigr)+ 123456.$ Compute the integer $\mathcal{K}$ .\

Ground Truth

124048

Execution Trace

Reasoning Process

[THOUGHT] We need to solve a multi-step problem. First, part I: given f(t) = cos(at) for |t| ≤ π/(2a), zero otherwise. Compute:

Fourier transform F(ω) = ∫{-∞}^{∞} f(t) e^{-iωt} dt = ∫{-π/(2a)}^{π/(2a)} cos(at) e^{-iωt} dt.

We can compute analytically using sympy.

Autocorrelation R_f(τ) = ∫{-∞}^{∞} f(t) f(t-τ) dt. Since f is even? cos(at) is even, and support symmetric. So R_f(τ) = ∫{-∞}^{∞} f(t) f(t-τ) dt. We can compute by convolution: R_f(τ) = (f ⋆ f)(τ) where ⋆ denotes cross-correlation: (f ⋆ f)(τ) = ∫ f(t) f(t-τ) dt. For real even f, autocorrelation equals convolution f * f(-τ)? Actually for real f, autocorrelation defined as ∫ f(t) f(t-τ) dt. For even f, it's same as convolution f * f(τ) maybe with a sign? Let's compute directly using sympy.
Fourier transform of R_f(τ): ℱ[R_f(τ)] = ? By Wiener–Khinchin theorem, Fourier transform of autocorrelation is |F(ω)|^2. So ℱ[R_f(τ)] = |F(ω)|^2. But we can also compute directly.

Then part II: evaluate integrals using results from I.

Given f(t) = cos(at) for |t| ≤ π/(2a). Let a be general. Then we can compute F(ω). Then maybe set a appropriately to match integrals.

The integrals are:

(1) ∫_{-∞}^{∞} cos²(π x / 2) / (x² - 1)² dx.

(2) ∫_{-∞}^{∞} cos⁴(π x / 2) / (x² - 1)⁴ dx.

We need to relate these to Fourier transforms or autocorrelations.

Observe that cos²(π x / 2) = (1 + cos(π x))/2. But denominator (x² - 1)² = (x-1)² (x+1)². Might be related to F(ω) squared? Possibly using Parseval's identity: ∫ |F(ω)|² dω = 2π ∫ |f(t)|² dt. But we have integrals of rational functions times cos powers.

Alternatively, note that F(ω) for f(t) = cos(at) windowed might be something like sinc-like. Possibly the integrals can be expressed as integrals of |F(ω)|² or |F(ω)|⁴ etc.

Let's compute F(ω) explicitly.

We'll use sympy to compute F(ω) = ∫_{-π/(2a)}^{π/(2a)} cos(at) e^{-iωt} dt.

Let's do step by step in Python.

[/THOUGHT] [PYTHON] import sympy as sp

Define symbols

a, t, omega = sp.symbols('a t omega', real=True, positive=True)

Fourier transform integral

F = sp.integrate(sp.cos(at) * sp.exp(-sp.Iomegat), (t, -sp.pi/(2a), sp.pi/(2*a))) sp.simplify(F) [/PYTHON]

Tool Output

[PYTHON OUTPUT] Piecewise((2*a*cos(pi*omega/(2*a))/(a**2 - omega**2), Ne(a, omega)), ((2*a*exp(I*pi*omega/a)*sin(pi*omega/(2*a)) + 2*a*sin(pi*omega/(2*a)) + 2*pi*omega*exp(I*pi*omega/(2*a)))*exp(-I*pi*omega/(2*a))/(4*a*omega), True)) [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The output gives F(ω) = 2acos(πω/(2a))/(a^2 - ω^2) for a ≠ ω. For ω = a, there's a separate expression. But we can simplify further: The piecewise can be combined. Actually, we can compute F(ω) manually: ∫ cos(at) e^{-iωt} dt = ∫ cos(at) cos(ωt) dt - i ∫ cos(at) sin(ωt) dt. Since cos(at) is even, the imaginary part integrates to zero because sin(ωt) is odd. So F(ω) is real and even. Indeed, the expression 2acos(πω/(2a))/(a^2 - ω^2) is real and even. For ω = a, we need limit. But for our purposes, we might not need ω = a exactly.

Now compute autocorrelation R_f(τ) = ∫ f(t) f(t-τ) dt. f(t) is even, so R_f(τ) is even. We can compute via convolution: R_f(τ) = ∫_{-∞}^{∞} f(t) f(t-τ) dt. Since f is zero outside |t| ≤ π/(2a), we need integration limits where both f(t) and f(t-τ) nonzero. That is |t| ≤ π/(2a) and |t-τ| ≤ π/(2a). This gives t ∈ [max(-π/(2a), τ - π/(2a)), min(π/(2a), τ + π/(2a))] intersection. We can compute piecewise.

But maybe easier: Use Wiener–Khinchin: ℱ[R_f(τ)] = |F(ω)|^2. Since F(ω) is real, |F(ω)|^2 = F(ω)^2. So R_f(τ) = ℱ^{-1}[F(ω)^2]. That might be easier to compute inverse Fourier transform of F(ω)^2.

But we need explicit R_f(τ) for part I(2). Let's compute directly using sympy.

Also, for part II, we need to evaluate integrals involving cos^2(π x/2)/(x^2-1)^2 etc. Notice that F(ω) for a = ? Possibly set a = π/2? Actually f(t) = cos(at) windowed. The integrals have cos(π x/2). Maybe if we set a = π/2? Let's see: F(ω) = 2acos(π ω/(2a))/(a^2 - ω^2). If we set a = π/2, then F(ω) = πcos(ω)/( (π/2)^2 - ω^2 ) = π*cos(ω)/( π^2/4 - ω^2 ). That doesn't directly match denominator (x^2-1)^2.

Wait, integrals are over x, which is like ω variable? Possibly the integrals are related to ∫ |F(ω)|^2 dω or ∫ |F(ω)|^4 dω? Because ℱ[R_f(τ)] = |F(ω)|^2. And Parseval: ∫ |F(ω)|^2 dω = 2π ∫ |f(t)|^2 dt. But integrals II have denominator (x^2-1)^2 and cos^2(π x/2). Maybe they are ∫ F(ω)^2 dω? Not exactly.

Let's compute R_f(τ) explicitly for general a, then its Fourier transform should be F(ω)^2. Then maybe we can relate integrals II to values of R_f(τ) at certain τ? Or to integrals of F(ω)^2 times something.

Alternatively, note that ∫_{-∞}^{∞} cos^2(π x/2)/(x^2-1)^2 dx = ? Possibly use contour integration. But the problem suggests using results from I. So likely we need to choose a appropriately so that F(ω) or R_f(τ) matches integrand.

Let's compute F(ω) for a = 1: F(ω) = 2*cos(π ω/2)/(1 - ω^2). Then F(ω)^2 = 4 cos^2(π ω/2)/(1 - ω^2)^2. That's exactly 4 times the integrand of II(1) if we replace ω with x. Indeed, ∫_{-∞}^{∞} cos^2(π ω/2)/(1 - ω^2)^2 dω = (1/4) ∫ F(ω)^2 dω. But ∫ F(ω)^2 dω = ∫ |F(ω)|^2 dω because F is real. And by Parseval, ∫ |F(ω)|^2 dω = 2π ∫ |f(t)|^2 dt. So we can compute ∫ f(t)^2 dt for a=1.

Thus A = ∫_{-∞}^{∞} cos^2(π x/2)/(x^2-1)^2 dx = (1/4) ∫ F(ω)^2 dω = (1/4) * (2π ∫ f(t)^2 dt) = (π/2) ∫ f(t)^2 dt.

Similarly for II(2): cos^4(π x/2)/(x^2-1)^4. Note that F(ω)^4 = 16 cos^4(π ω/2)/(1 - ω^2)^4. So ∫ cos^4(π ω/2)/(1 - ω^2)^4 dω = (1/16) ∫ F(ω)^4 dω. But ∫ F(ω)^4 dω = ∫ |F(ω)|^4 dω. However Parseval relates ∫ |F(ω)|^2 dω to ∫ |f(t)|^2 dt, not directly to ∫ |F(ω)|^4 dω. However, note that ℱ[R_f(τ)] = F(ω)^2. So ℱ[R_f(τ)^2]? Actually, convolution theorem: ℱ[(f * f)(τ)] = F(ω)^2, where * is convolution. But autocorrelation is not convolution exactly but cross-correlation. For even real f, autocorrelation equals convolution f * f. Indeed, R_f(τ) = ∫ f(t) f(t-τ) dt = (f * f)(τ) if f is even? Actually convolution: (f * f)(τ) = ∫ f(t) f(τ - t) dt. For even f, f(τ - t) = f(t - τ). So convolution equals cross-correlation. So R_f = f * f. Then ℱ[R_f] = F^2. Then ℱ[R_f * R_f] = ℱ[R_f]^2 = F^4. And ∫ F(ω)^4 dω = 2π ∫ R_f(τ)^2 dτ by Parseval applied to R_f. So we can compute B = ∫ cos^4(π x/2)/(x^2-1)^4 dx = (1/16) ∫ F(ω)^4 dω = (1/16) * (2π ∫ R_f(τ)^2 dτ) = (π/8) ∫ R_f(τ)^2 dτ.

Thus we need to compute for a=1: ∫ f(t)^2 dt and ∫ R_f(τ)^2 dτ.

Let's compute these integrals using sympy.

First, f(t) = cos(t) for |t| ≤ π/2, zero elsewhere. So ∫ f(t)^2 dt = ∫_{-π/2}^{π/2} cos^2(t) dt.

Second, compute R_f(τ) for a=1. Then compute ∫ R_f(τ)^2 dτ.

Let's do step by step.

[/THOUGHT] [PYTHON] import sympy as sp

Define symbols

t, tau = sp.symbols('t tau', real=True) a = 1

Define f(t)

f = sp.Piecewise((sp.cos(at), sp.Abs(t) <= sp.pi/(2a)), (0, True))

Compute integral of f(t)^2

int_f_sq = sp.integrate(f**2, (t, -sp.oo, sp.oo)) print("∫ f(t)^2 dt =", int_f_sq.simplify())

Compute autocorrelation R_f(tau) = ∫ f(t) f(t - tau) dt

R = sp.integrate(f.subs(t, t) * f.subs(t, t - tau), (t, -sp.oo, sp.oo)) print("R_f(τ) =", R.simplify())

Compute integral of R_f(τ)^2

int_R_sq = sp.integrate(R**2, (tau, -sp.oo, sp.oo)) print("∫ R_f(τ)^2 dτ =", int_R_sq.simplify()) [/PYTHON]

Tool Output

[PYTHON OUTPUT] ∫ f(t)^2 dt = pi/2 R_f(τ) = sin(tau - 2*Max(-pi/2, tau - pi/2))/4 - sin(tau - 2*Max(-pi/2, tau - pi/2, Min(pi/2, tau + pi/2)))/4 - cos(tau)*Max(-pi/2, tau - pi/2)/2 + cos(tau)*Max(-pi/2, tau - pi/2, Min(pi/2, tau + pi/2))/2 ∫ R_f(τ)^2 dτ = pi*(15 + 2*pi**2)/24 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] We have: ∫ f(t)^2 dt = π/2. Thus A = ∫_{-∞}^{∞} cos²(π x/2)/(x²-1)² dx = (π/2) * ∫ f(t)^2 dt = (π/2) * (π/2) = π²/4? Wait earlier: A = (π/2) ∫ f(t)^2 dt = (π/2)*(π/2) = π²/4.

But we need to verify: earlier we had A = (π/2) ∫ f(t)^2 dt? Actually we derived A = (π/2) ∫ f(t)^2 dt? Let's re-check: ∫ cos²(π ω/2)/(1 - ω²)² dω = (1/4) ∫ F(ω)² dω. And ∫ F(ω)² dω = 2π ∫ f(t)² dt. So (1/4)2π ∫ f(t)² dt = (π/2) ∫ f(t)² dt. Yes. So A = (π/2)(π/2) = π²/4.

Now B = ∫ cos⁴(π x/2)/(x²-1)⁴ dx = (π/8) ∫ R_f(τ)² dτ. We computed ∫ R_f(τ)² dτ = π*(15 + 2π²)/24. So B = (π/8) * π*(15 + 2π²)/24 = π²*(15 + 2π²)/(192).

Now compute F(0) for a=1: F(ω) = 2cos(π ω/2)/(1 - ω²). At ω=0, F(0)=2cos(0)/(1-0)=2.

Now compute 𝒦 = 37*(F(0))^4 + 999*(96B - 16A² - 30A) + 123456.

Compute step by step:

F(0)^4 = 2^4 = 16. So 37*16 = 592.

A = π²/4. A² = π⁴/16. 30A = 30π²/4 = (15π²)/2.

96B = 96 * π²*(15 + 2π²)/192 = (96/192) * π²*(15 + 2π²) = (1/2) * π²*(15 + 2π²) = (π²*(15 + 2π²))/2.

Thus 96B - 16A² - 30A = [π²*(15 + 2π²)/2] - [16*(π⁴/16)] - [15π²/2] = (π²*(15 + 2π²)/2) - π⁴ - (15π²/2) = combine first and third: (π²*(15 + 2π²) - 15π²)/2 - π⁴ = (π²*15 + 2π⁴ - 15π²)/2 - π⁴ = (2π⁴)/2 - π⁴ = π⁴ - π⁴ = 0.

So 96B - 16A² - 30A = 0.

Thus 𝒦 = 3716 + 9990 + 123456 = 592 + 123456 = 124048.

Now compute numerically to ensure no rounding errors.

Let's compute with Python to verify.

[/THOUGHT] [PYTHON] import sympy as sp pi = sp.pi A = pi2/4 B = pi2*(15 + 2pi2)/192 F0 = 2 term1 = 37 * F04 term2 = 999 * (96B - 16A**2 - 30A) K = term1 + term2 + 123456 print("A =", A) print("B =", B) print("96B - 16A^2 - 30A =", (96B - 16A**2 - 30*A).simplify()) print("K =", K) print("K numeric =", K.evalf()) [/PYTHON]

Tool Output

[PYTHON OUTPUT] A = pi**2/4 B = pi**2*(15 + 2*pi**2)/192 96B - 16A^2 - 30A = 0 K = -999*pi**4 - 14985*pi**2/2 + 124048 + 999*pi**2*(15 + 2*pi**2)/2 K numeric = 124048.000000000 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python output confirms that 96B - 16A² - 30A = 0, and 𝒦 = 124048 exactly. Therefore the integer 𝒦 is 124048.

[/THOUGHT] \boxed{124048}

Final Model Output

124048

Comparative Analysis (P4) (Average across runs)

Avg Time Taken

GPT-5.1 (high)

414.5

Gemini-3-Pro-Preview

235.9

Deepseek-V3.2

172.0

Solar-Pro-2 (31B)(high)

168.4

Llama-VARCO-8B-Instruct

128.4

A.X-4.0 (72B)

124.6

Grok-4.1-fast

63.1

Claude-Opus-4.5

52.5

HCX-007(high)

32.4

EXAONE-4.0.1-32B (high)

9.1

Avg Token Usage

Solar-Pro-2 (31B)(high)

22565.0

GPT-5.1 (high)

18561.0

Deepseek-V3.2

17612.0

Grok-4.1-fast

13781.0

Gemini-3-Pro-Preview

12183.0

Claude-Opus-4.5

9408.0

A.X-4.0 (72B)

9126.0

Llama-VARCO-8B-Instruct

8388.0

EXAONE-4.0.1-32B (high)

5657.0

HCX-007(high)

5162.0